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Eduardwww [97]
4 years ago
9

A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of

300 KNm, as shown. The modulus of rigidity G for the steel is 80 GPa. Determine: (a) The maximum shearing stress in the shaft. (b) The shearing stress on a traverse cross section at the inside surface of the shaft (c) The magnitude of the angle of twist for a (L) -2.5 m length.
Engineering
1 answer:
-Dominant- [34]4 years ago
8 0

Answer:

a.  \tau=51.55 MPa

b.\tau=42.95MPa

c.\theta=7.67\times 10^{-3} rad.

Explanation:

Given: D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa

We know that

\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}

J for hollow shaft J=\dfrac{\pi (D_o^4-D_i^4)}{64}

(a)

 Maximum shear stress \tau =\dfrac{16T}{\pi Do^3(1-K^4)}

      K=\dfrac{D_i}{D_o}⇒K=0.83

\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}

   \tau=51.55 MPa

(b)

We know that \tau \alpha r

So \dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}

\dfrac{51.55}{\tau}=\dfrac{210}{175}

\tau=42.95MPa

(c)

\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}

\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}

\theta=7.67\times 10^{-3} rad.

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